Answer:
nothing will happen the cart will be broken or as it is
The kayaker has velocity vector
<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )
<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )
and the current has velocity vector
<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )
<em>w</em> ≈ (0.884 m/s) (<em>i</em> - <em>j</em> )
The kayaker's total velocity is the sum of these:
<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>
That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that
tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s) → <em>θ</em> ≈ 18.4º
or about 18.4º north of east.
Answer:
d. 3332.5 [N]
Explanation:
To solve this problem we will use newton's second law, which tells us that the sum of forces is equal to the product of mass by acceleration.
Here we have two forces, the force that pushes the car to move forward and the friction force.
The friction force is equal to the product of the normal force by the coefficient of friction.
f = N * μ
f = (m*g) * μ
where:
N = weight of the car = 2150*9.81 = 21091.5 [N]
μ = 0.25
f = (21091.5) * 0.25
f = 5273 [N]
Now as the car is moving forward, the car wheels move clockwise. The friction force between the wheels of the car and the pavement must be counterclockwise, i.e. counterclockwise. Therefore the direction of this force is forward. This way we have:
F + f = m*a
F + 5273 = 2150*4
F = 8600 - 5273
F = 3327 [N]
Therefore the answer is d.
Https://www.google.com/search?q=Why+do+atoms+have+an+overall+neutral+charge%3F&oq=Why+do+atoms+have+...
Neutrons are no exception. So, if an atom<span> has equal numbers of electrons and protons, the </span>charges<span> cancel each other out and the </span>atom<span> has a </span>neutral charge<span>.
</span>
Answer: option D. the ratio of the population of male deer is not constant.
Explanation:
The bar graph permits to compare the results for two different populations: male and female deer in a very easy visual way.
These features are remarkable:
- The polulation of male deer (blue bars) decrease from 1961 to 1971, then increase in the next 10 year, decrease in the next decade, and increase for the next two decades. So, its trend is erratic, with ups and downs.
This discards the option A, which states that the population of male deer increases each decade from 1961 to 2011.
- The population of female deer (purple or brown bars) decreases every decade.
This discards the option B. which states that when the polulation of male deer increases, the poluplation of female deer also increases.
- The populations never are equal, hence this discards the option C.
- Since, one popultion increases and decreases, while the other population only decreases, you conclude that the ratio of the population of male deer to female deer is not constant, which is the option D.
